Question

For the given aqueous reaction which of the statement(s) is/are true?
$\text{Excess}KI+K_3\left[Fe\right(CN{)}_6]\stackrel{Dil.H_{2}S{O}_4}{\to }\text{Brownish yellow solution}\downarrow ZnS{O}_4\text{(White precipitate + Brownish yellow filtrate)}\downarrow N{a}_2{S}_2{O}_3\text{Colourless solution}$

• A. The first reaction is a redox reaction
• B. White precipitate is $Z{n}_3\left[Fe\right(CN{)}_6{]}_2$
• C. Addition of filtrate to starch solution gives blue colour
• D. White precipitate is soluble in NaOH solution

Answer 1

Answer: (A), (C), (D)

White precipitate is soluble in NaOH solution

$K_3\left[Fe\right(CN{)}_6]+KI\text{(excess)}\to K_4[Fe\left(CN{)}_6\right]+\underset{\text{Brownish yellow solution}}{K{I}_3\left(redox\right)}{K}_4\left[Fe\right(CN{)}_6]+K{I}_3+ZnS{O}_4\to \underset{\text{White ppt}}{K_{2}Zn\left[Fe\right(CN{)}_6]}+\underset{\text{Brownish yellow filtrate}}{K{I}_3}+K_{2}S{O}_4\underset{\text{Brownish yellow filtrate}}{I_3^{-}}+2N{a}_2{S}_2{O}_3\to \underset{Clearsolution}{N{a}_2{S}_4{O}_6}+2NaI+\underset{\text{Turns starch solution blue}}{I_2}{K}_{2}Zn[Fe\left(CN{)}_6\right]\text{reacts with NaOH as}{K}_{2}Zn\left[Fe\right(CN{)}_6]+NaOH\to [Zn(OH{)}_4{]}^{2-}+[Fe(CN{)}_6{]}^{4-}$